In a laboratory,two bottles contain a mixture | vedclass

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(C) Let the ratio of the first mixture to the second mixture be $x:y$.In the first bottle,the fraction of acid is $\frac{2}{2+5} = \frac{2}{7}$.In the

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$21:8$

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(C) Let the ratio of the first mixture to the second mixture be $x:y$.In the first bottle,the fraction of acid is $\frac{2}{2+5} = \frac{2}{7}$.In the second bottle,the fraction of acid is $\frac{7}{7+3} = \frac{7}{10}$.In the final mixture,the fraction of acid is $\frac{2}{2+3} = \frac{2}{5}$.Using the rule of alligation:Difference between the second mixture and the final mixture: $\frac{7}{10} - \frac{2}{5} = \frac{7-4}{10} = \frac{3}{10}$.Difference between the final mixture and the first mixture: $\frac{2}{5} - \frac{2}{7} = \frac{14-10}{35} = \frac{4}{35}$.The ratio $x:y$ is $\frac{3}{10} : \frac{4}{35}$.Multiplying both sides by $70$,we get $x:y = (\frac{3}{10} 	imes 70) : (\frac{4}{35} 	imes 70) = 21 : 8$.

In a laboratory,two bottles contain a mixture of acid and water in the ratio $2:5$ in the first bottle and $7:3$ in the second. The ratio in which the contents of these two bottles should be mixed such that the new mixture has acid and water in the ratio $2:3$ is

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